# Find subtended angle

• Mar 9th 2010, 07:38 PM
arze
Find subtended angle
A triangle ABC is inscribed in a circle with center O. The line CD is perpendicular to the plane of the circle and of length CD=10cm. If angle DAC=$\displaystyle 68^o12'$, the angle DBC=$\displaystyle 60^o19'$ and AB=6.67cm calculate the angle subtended by each of the sides of the triangle ABC at the center O.

I calculated AC=4.00cm and BC=5.70cm. I don't know what to do next. I've tried using the cosine rule to find in terms of r, the radius of the circle, the angles, then find the value of r from the area.
$\displaystyle \cos AOC=\frac{2r^2-4^2}{2r^2}$
$\displaystyle \cos BOC=\frac{2r^2-5.7^2}{2r^2}$
$\displaystyle \cos AOB=\frac{2r^2-6.67^2}{2r^2}$
Area=$\displaystyle \frac{1}{2}r(6.67+5.70+4.00)=8.18r$
Area=$\displaystyle \frac{1}{2}\times 6.67\times 4.00\times\sin\frac{BOC}{2}=13.34\sin\frac{BOC}{2}$
so then r=$\displaystyle \frac{13.34\sin\frac{BOC}{2}}{8.18}=1.631\sin\frac {BOC}{2}$
Then i filled in r into $\displaystyle \cos BOC=\frac{2r^2-5.7^2}{2r^2}$ and got, where $\displaystyle \frac{BOC}{2}=\theta$
$\displaystyle \cos 2\theta=1-\frac{6.109}{\sin^2\theta}$
I solve this equation for $\displaystyle \theta$ but the values of $\displaystyle \cos 2\theta$ were greater than 1 so this is the wrong approach.
Thanks
• Mar 9th 2010, 08:43 PM
Soroban
Hello, arze!

You were that close to the solution . . .

Quote:

A triangle $\displaystyle ABC$ is inscribed in a circle with center $\displaystyle O.$
The line $\displaystyle CD$ is perpendicular to the plane of the circle and of length $\displaystyle CD=10$ cm.
If $\displaystyle \angle DAC \,=\, 68^o12',\;\angle DBC\,=\,60^o19',\;AB\,=\,6.67$ cm,
calculate the angle subtended by each of the sides of the triangle $\displaystyle ABC$ at center $\displaystyle O.$

I calculated AC=4.00 cm and BC=5.70 cm. . . . . Good!

We have this triangle:
Code:

            C             o           *  *         4 *    *  5.7         *        *         *          *       *              *     A o  *  *  *  *  *  o B               6.67
Find the angles of this triangle.

$\displaystyle \cos A \:=\:\frac{6.67^2 + 4^2 - 5.7^2}{2(6.67)(4)} \:=\:0.524717016 \quad\Rightarrow\quad \angle A \:\approx\:58.3^o$

$\displaystyle \cos B \:=\:\frac{6.67^2 + 5.7^2 - 4^2}{2(6.67)(5.7)} \:=\:0.801952971 \quad\Rightarrow\quad \angle B \:\approx\:36.7^o$

$\displaystyle \cos C \:=\:\frac{5.7^2+4^2 - 6.67^2}{2(5.7)(4)} \:=\:0.087743421 \quad\Rightarrow\quad \angle C \:\approx\:85.0^o$

Since $\displaystyle \angle A, \angle B, \angle C$ are inscribed angles, we know their corresponding arcs.

. . $\displaystyle \begin{array}{ccccccc} \angle A &=& 58.3 & \Rightarrow & \text{arc}(BC) &=& 116.6^o \\ \angle B &=& 36.7^o & \Rightarrow & \text{arc}(AC) &=& 73.4^o \\ \angle C &=& 85.0^o & \Rightarrow & \text{arc}(AB) &=& 170.0^o \end{array}$

Since $\displaystyle \angle AOB, \angle BOC, \angle AOC$ are central angles, we're done!

. . $\displaystyle \begin{array}{ccccccc}\text{arc}(AB) &=& 170.0^2 & \Rightarrow & \angle AOB &=& 170.0^o \\ \text{arc}(BC) &=& 116.6^o & \Rightarrow & \angle BOC &=& 116.6^o \\ \text{arc}(AC) &=& 73.4^o & \Rightarrow & \angle AOC &=& 73.4^o \end{array}$